# Analytic functions where all derivatives vanish at infinity and which are bounded

Let $$C_0(\mathbb{R})$$ denote the analytic functions $$f : \mathbb{R} \rightarrow \mathbb{R}$$.

I wonder whether there a functions $$f \in C_0(\mathbb{R})$$ with $$f \neq 0$$, such that there is a constant $$C$$, with $$\left| \frac{d^kf}{d^kx} \right | \leq C$$ for all $$k \geq 0$$, and $$\frac{d^kf}{dx}$$ vanish both at $$- \infty$$ and $$\infty$$ for all $$k \geq 0$$.

• What about $(\sin x)/x$? Oct 4, 2020 at 17:29
• also, technically, since you didn't specify non-trivial, I guess $f\equiv 0$ is another exampl. Oct 4, 2020 at 17:37
• @Willie Wong Thanks for pointing it out, I amended it. Oct 4, 2020 at 17:54
• @RichardStanley It seems $sin(x)/x$ is doing the job. However, is there an easy proof for it? Oct 4, 2020 at 21:10
• @tobias: for a rescaled sinc function you can also get the proof using my answer. the fact that its Fourier transform is the rectangle function and not smooth is inconsequential: the Fourier transform of all its derivatives are bounded with compact support, so the decay at infinity can be also gotten via Riemann-Lebesgue. Oct 5, 2020 at 1:55

Yes.

Let $$\phi$$ be any smooth function with compact support on the interval $$[-1,1]$$.

Set $$f$$ to be the inverse Fourier transform of $$\phi$$.

Since $$\phi$$ is in Schwartz class, so is $$f$$, and all of its derivatives decay to zero as one approach $$\pm\infty$$.

You can estimate

$$|f^{(k)}(x) | \lesssim \| |\xi|^k \phi(\xi) \|_{L^1} \leq 2 \|\phi\|_{L^\infty} =: C$$

$$f$$ is analytic by Paley-Wiener.

• Why do you specify “inverse” Fourier transform and not simply the Fourier transform? For thepurposes of question it amounts to the same thing. Oct 4, 2020 at 18:11
• Because for my own sanity I like to keep track of which functions are on the frequency domain and which are on the time domain. (Force of habit.) Oct 4, 2020 at 18:45